Torque and Angular Momentum
Key idea: Understand angular momentum and torque, use τ = dL/dt for rotation, and apply conservation of angular momentum when external torque is negligible.
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The core idea
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Learning objectives
- show an understanding of and use the terms angular displacement, angular velocity, and angular acceleration of a rigid body with respect to a fixed axis
- solve problems using the equations of motion for uniform angular acceleration that are analogous to the equations of motion for uniform linear acceleration
- show an understanding of and use the terms angular momentum and moment of inertia of a rotating rigid body
- calculate the moment of inertia about an axis for simple bodies by using calculus, the parallel-axis theorem or otherwise (knowledge of the perpendicular-axis theorem and mathematical derivation of the moment of inertia for spheres are not required)
- show an understanding of torque produced by a force relative to a reference point, and apply the principle that torque is related to the rate of change of angular momentum to solve problems, such as those involving point masses, rigid bodies, or bodies with a variable moment of inertia e.g. an ice-skater
- Apply Eₖ,rot = ½Iω² to the rotational kinetic energy of a rigid body.
- Derive Eₖ,rot = ½Iω² for a rigid body from the equations of motion.
- recall and apply the result that the motion of a rigid body can be regarded as translational motion of its centre of mass with rotational motion about an axis through the centre of mass to solve problems, including the use of F ⩽ µN for solid surfaces in no-slip contact (no distinction is made between the coefficient of static and kinetic friction)
Angular momentum is the rotational analogue of linear momentum. The central H3 idea is: ∑τₑₓₜ = dL/dt for components about a fixed axis. For a rigid body with constant I, this reduces to ∑τₑₓₜ = Iα. Here, “net torque” always means the resultant external torque about the stated axis.
1. Definitions (Must Know)
- Angular momentum, L:
- particle about an origin: vector L = vector r × vector p
- fixed-axis rigid body (magnitude): L = Iω
- Torque, τ about an axis/pivot:
- magnitude: τ = rF sin φ = Fd_⊥
- Torque–angular momentum relation: ∑τₑₓₜ = dL/dt
- Symbols used in this lesson: vector r position vector (m), vector p momentum (kg m s⁻¹), vector L angular momentum (kg m² s⁻¹), τ torque (N m), I moment of inertia (kg m²), ω angular speed (rad s⁻¹), α angular acceleration (rad s⁻²).
Quick comparison:
| System | Angular momentum | When to use |
|---|---|---|
| Particle about an origin | vector L = vector r × vector p | Point-mass / orbit-type setups |
| Fixed-axis rigid body | L = Iω | Most H3 rigid-body rotation questions |
2. Key Ideas (What Earns Marks)
- Always specify the reference point/axis when talking about torque or angular momentum.
- Torque changes angular momentum: no net external torque ⇒ angular momentum is conserved (about that axis).
- For fixed-axis rigid bodies: L = Iω is the high-speed exam tool.
- Variable I problems (ice skater): if τₑₓₜ ≈ 0, then Iω stays constant even though ω changes.
3. Detailed Explanations
A. Why the axis matters
Both τ and L are defined about a point/axis. Changing the axis changes:
- the moment arm d_⊥ (hence the torque), and
- the distance vector vector r (hence the angular momentum).
B. Linking torque and angular momentum (fixed-axis intuition)
For a point mass moving in a circle of radius r about an axis, the tangential force Fₜ changes the tangential speed, hence changes the “rotational motion”. The torque about the axis is: τ = rFₜ
For a rigid body about a fixed axis, all the parts rotate together with the same ω. The result you use is: ∑τₑₓₜ = dL/dt
If L = Iω and I is constant:
This connects directly to the previous two lessons:
C. When I changes (ice skater idea)
If there is negligible external torque about the spin axis, then dL/dt ≈ 0, so L is approximately constant.
For a spinning skater, you can model L ≈ Iω. Pulling arms in reduces I, so ω increases to keep Iω constant.
4. Common Mistakes
- Writing “angular momentum is conserved” without stating “when external torque is negligible (about the chosen axis)”.
- Using L = Iω for a particle not in rigid rotation about the axis (use vector L = vector r × vector p if needed).
- Forgetting that τ = Fd_⊥ depends on the perpendicular distance to the line of action (not just “the radius”).
- Mixing up L (angular momentum) with I (moment of inertia) because both appear in rotational formulas.
5. Exam Tips
- If the question screams “conservation”, check the external torque about the relevant axis:
- Is there a pivot reaction through the axis? (often gives zero torque)
- Is friction/air resistance negligible? (often assumed)
- Use the reasoning chain: negligible external torque ⇒ constant L ⇒ constant Iω for fixed-axis rotation.
- For fixed-axis questions, start from L = Iω and ∑τₑₓₜ = dL/dt; it’s faster than re-deriving.
6. Worked Examples
Modelled example 1
Net torque gives rate of change of angular momentum
Problem
Study the worked solution
Use angular impulse
Method
Δ L = τₙₑₜΔ t.Reason
Integrating dL/dt = τₙₑₜ over constant torque gives the product.Working
Δ L = ∫τ dtEvaluate
Method
|Δ L| = 6.0 kg m² s⁻¹.Reason
Multiply torque by duration.Working
(1.5)(4.0) = 6.0 kg m² s⁻¹
Guided practice 2
Ice skater (variable moment of inertia)
Problem
Try this before viewing the solution
Hints
Hint 1: check the external torque
View solution step by step
Conserve angular momentum
Method
I₁ω₁ = I₂ω₂.Reason
τₑₓₜ ≈ 0.Working
L₁ = L₂Solve
Method
ω₂ = 5.0 rad s⁻¹.Reason
Reducing inertia requires increased angular speed for fixed L.Working
(3.0)(2.0)/1.2 = 5.0
Common misconception 3
Fixed-axis rigid body: torque and angular acceleration
Learner claim
Try this before viewing the solution
View solution step by step
Find angular acceleration
Method
α = 4.0 rad s⁻².Reason
For constant I, τ = Iα.Working
α = 2.4/0.60 = 4.0Find momentum rate
Method
dL/dt = 2.4 kg m² s⁻².Reason
The general law is τₙₑₜ = dL/dt.Working
dL/dt = τₙₑₜ = 2.4
Examiner practice 4
Finding L from I and ω
Examination question
Try this before viewing the solution
View solution step by step
State relation
1 markMethod
L = Iω.Reason
The body rotates rigidly about the stated fixed axis.Working
L = IωSubstitute
1 markMethod
L = (0.40)(12).Reason
Use SI quantities.Working
L = 4.8Report
1 markMethod
L = 4.8 kg m² s⁻¹.Reason
These are angular-momentum units.Working
4.8 kg m² s⁻¹
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark relation, substitution and result.
Challenge 5
Angular impulse from a changing torque (area-under-graph idea)
Independent transfer
Try this before viewing the solution
Hints
Hint 1: use area under torque-time graph
View solution step by step
Set up the integral
Method
Δ L = ∫₀² 3.0t dt.Reason
Torque is the time derivative of angular momentum.Working
Δ L = ∫τ dtEvaluate
Method
Δ L = 6.0 kg m² s⁻¹.Reason
The integral equals the triangular area under the torque-time line.Working
3.0[t²/2]₀² = 6.0
7. Mind Stretchers
Mind stretcher 1: Why a figure skater speeds up (concept + energy)Extension
If I decreases and L is conserved, ω increases. Where does the extra rotational kinetic energy come from?
Answer
From work done by the skater’s muscles while pulling the arms inward (internal forces can do work on parts of the body, increasing rotational kinetic energy even though external torque is negligible).
Mind stretcher 2: “Conserved about which axis?”Extension
An object is rotating while a force acts on it. How do you decide whether angular momentum is conserved in an exam question?
Answer
Pick the axis or reference point, then check the external torque about it. If τₑₓₜ ≈ 0, angular momentum is conserved about that axis. If there is a significant external torque about your chosen axis, angular momentum about that axis is not conserved.
8. Optional/Enrichment: Vector Form and 3D Rotation
The full relation is vector-based: ∑ vector τₑₓₜ = (d vector L)/dt with vector L = vector r × vector p. For 3D rigid body rotation (changing axis direction), you need more advanced dynamics; H3 fixed-axis questions typically avoid this unless explicitly guided.
Next step
Return to the Rotational Motion hub, or continue in sequence to Rotational Kinetic Energy.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H3 Physics
- Edition
- GCE A-Level H3 Physics 2027